Evaluating \(\displaystyle{\prod_{{{i}={1}}}^{{{60}}}}{\sin{{\left({3}{i}-{2}\right)}}}{\sin{{\left({3}{i}-{1}\right)}}}\)

siliciooy0j

siliciooy0j

Answered question

2022-04-02

Evaluating i=160sin(3i2)sin(3i1)

Answer & Explanation

kanonickiuoeh

kanonickiuoeh

Beginner2022-04-03Added 8 answers

In conventional notation, you are after
A=i=160sin(3i2)π180sin(3i1)π180
Then
A=P180P60
where
PN=j=1N1sinjπN
It is well-known that
PN=N2N1
Thus
A=32120
A hint for (1). Consider the product
N=j=1N1(1exp(2πijN))

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